គេទំលាក់គ្រាប់ឃ្លីពីរគ្រាប់ A និង B ពីទីតាំងតែមួយ តែនៅខណៈពេលខុសគ្នា។ ក្រោយ 2s គិតចាប់ពី ពេលដែលគ្រាប់ឃ្លីទីពីរ B បានធ្លាក់ នោះគេសង្កេតឃើញថា ចំងាយរវាងគ្រាប់ឃ្លីទាំងពីរស្មើ 60m។ តើគ្រាប់ឃ្លី B ត្រូវបានគេទំលាក់យឺតជាងគ្រាប់ឃ្លី A រយៈពេលប៉ុន្មាន? (យក g=10 (SI)).
Solved by brother Soun Sovathana.
ReplyDeleteet t is duration that 2nd ball release after 1st ball
so t+2 is duration of 1st ball motion
by h=g(2)^2/2=10(4)/2=20m
then total distance of 1st ball is H=20+60=80m
and H=g(t+2)^2/2=80
=> t+2=4 , t=2s
Thus t=2s
tha roh post heuy :D lok bong post ouy bat :D
ReplyDeletethanks hehe
tha roh post, ey lov lok bong post ouy bat :D
ReplyDeletethanks hehe
Hjhj, just copy and past ^_^
DeleteTomorrow will have a great exercise, so don't forget to check it out na brother ^_^
DeleteThx for this exercise bro :D
ReplyDeleteYou are welcome ^_^
DeleteDon't forget to check this blog everyday na, cause i will post new exercise one a day.